Problem: Evaluate the following expression. Your answer must be exact. $(1+\sqrt{3}i)^6=$
Solution: The Strategy The easiest way to find $z^{n}$ for a complex number $z=({a}+{b}i)$ is using its modulus and argument. Therefore, our solution will consist of the following steps: Find the modulus and argument of $z$. [How is this done, in general?] Find the modulus and argument of $z^{n}$. [How is this done, in general?] Find the rectangular form $z^{n}$. Find the modulus and argument of $(1+\sqrt{3}i)$ $({1}+{\sqrt{3}}i)$ is of the form $({a}+{b}i)$, where ${a=1}$ and ${b=\sqrt{3}}$. Therefore: $\begin{aligned}r&=\sqrt{{a}^2 + {b}^2} \\\\&=\sqrt{({1})^2 + ({\sqrt{3}})^2} \\\\&=\sqrt{{1}+{3}} \\\\&=2\end{aligned}$ Using the arctangent formula, we have: $\begin{aligned}\theta&=\arctan\left(\dfrac{{b}}{{a}}\right) \\\\&=\arctan\left(\dfrac{{\sqrt{3}}}{{1}}\right) \\\\&=60^\circ\end{aligned}$ Since both ${a=1}$ and ${b=\sqrt{3}}$ are positive, $(1+\sqrt{3}i)$ lies in Quadrant $1$. Therefore, $\theta$ must be between $0^\circ$ and $90^\circ$, so our answer matches our requirements. Find the modulus and argument of $(1+\sqrt{3}i)^6$ We found that the modulus and argument of $({1}+{\sqrt{3}}i)$ are $2$ and $60^\circ$. Therefore, the modulus and argument of $({1}+{\sqrt{3}}i)^6$ are $2^6=64$ and $(60^\circ)\cdot6=360^\circ$. Find the rectangular form of $(1+\sqrt{3}i)^6$ Since the argument is $360°$, we know the number lies on the positive side of the real number axis and is therefore a positive real number. Since the modulus is $64$, our solution is $64$. [What does this look like graphically?] [How do we find this algebraically?] Summary $(1+\sqrt{3}i)^6=64$